problem solving involving linear equation

Word Problems on Linear Equations

Worked-out word problems on linear equations with solutions explained step-by-step in different types of examples.

There are several problems which involve relations among known and unknown numbers and can be put in the form of equations. The equations are generally stated in words and it is for this reason we refer to these problems as word problems. With the help of equations in one variable, we have already practiced equations to solve some real life problems.

Steps involved in solving a linear equation word problem: ● Read the problem carefully and note what is given and what is required and what is given. ● Denote the unknown by the variables as x, y, ……. ● Translate the problem to the language of mathematics or mathematical statements. ● Form the linear equation in one variable using the conditions given in the problems. ● Solve the equation for the unknown. ● Verify to be sure whether the answer satisfies the conditions of the problem.

Step-by-step application of linear equations to solve practical word problems:

1. The sum of two numbers is 25. One of the numbers exceeds the other by 9. Find the numbers. 

Solution: Then the other number = x + 9 Let the number be x.  Sum of two numbers = 25 According to question, x + x + 9 = 25 ⇒ 2x + 9 = 25 ⇒ 2x = 25 - 9 (transposing 9 to the R.H.S changes to -9)  ⇒ 2x = 16 ⇒ 2x/2 = 16/2 (divide by 2 on both the sides)  ⇒ x = 8 Therefore, x + 9 = 8 + 9 = 17 Therefore, the two numbers are 8 and 17.

2.The difference between the two numbers is 48. The ratio of the two numbers is 7:3. What are the two numbers?  Solution:   Let the common ratio be x.  Let the common ratio be x.  Their difference = 48 According to the question,  7x - 3x = 48  ⇒ 4x = 48  ⇒ x = 48/4  ⇒ x = 12 Therefore, 7x = 7 × 12 = 84           3x = 3 × 12 = 36  Therefore, the two numbers are 84 and 36.

3. The length of a rectangle is twice its breadth. If the perimeter is 72 metre, find the length and breadth of the rectangle.  Solution: Let the breadth of the rectangle be x,  Then the length of the rectangle = 2x Perimeter of the rectangle = 72 Therefore, according to the question 2(x + 2x) = 72 ⇒ 2 × 3x = 72 ⇒ 6x = 72  ⇒ x = 72/6 ⇒ x = 12 We know, length of the rectangle = 2x                       = 2 × 12 = 24 Therefore, length of the rectangle is 24 m and breadth of the rectangle is 12 m.

4. Aaron is 5 years younger than Ron. Four years later, Ron will be twice as old as Aaron. Find their present ages. 

Solution: Let Ron’s present age be x.  Then Aaron’s present age = x - 5 After 4 years Ron’s age = x + 4, Aaron’s age x - 5 + 4.  According to the question;  Ron will be twice as old as Aaron.  Therefore, x + 4 = 2(x - 5 + 4)  ⇒ x + 4 = 2(x - 1)  ⇒ x + 4 = 2x - 2 ⇒ x + 4 = 2x - 2 ⇒ x - 2x = -2 - 4 ⇒ -x = -6 ⇒ x = 6 Therefore, Aaron’s present age = x - 5 = 6 - 5 = 1 Therefore, present age of Ron = 6 years and present age of Aaron = 1 year.

5. A number is divided into two parts, such that one part is 10 more than the other. If the two parts are in the ratio 5 : 3, find the number and the two parts.  Solution: Let one part of the number be x Then the other part of the number = x + 10 The ratio of the two numbers is 5 : 3 Therefore, (x + 10)/x = 5/3 ⇒ 3(x + 10) = 5x  ⇒ 3x + 30 = 5x ⇒ 30 = 5x - 3x ⇒ 30 = 2x  ⇒ x = 30/2  ⇒ x = 15 Therefore, x + 10 = 15 + 10 = 25 Therefore, the number = 25 + 15 = 40  The two parts are 15 and 25. 

More solved examples with detailed explanation on the word problems on linear equations.

6. Robert’s father is 4 times as old as Robert. After 5 years, father will be three times as old as Robert. Find their present ages.  Solution: Let Robert’s age be x years.  Then Robert’s father’s age = 4x After 5 years, Robert’s age = x + 5 Father’s age = 4x + 5 According to the question,  4x + 5 = 3(x + 5)  ⇒ 4x + 5 = 3x + 15  ⇒ 4x - 3x = 15 - 5  ⇒ x = 10 ⇒ 4x = 4 × 10 = 40  Robert’s present age is 10 years and that of his father’s age = 40 years.  

7. The sum of two consecutive multiples of 5 is 55. Find these multiples.  Solution: Let the first multiple of 5 be x.  Then the other multiple of 5 will be x + 5 and their sum = 55 Therefore, x + x + 5 = 55 ⇒ 2x + 5 = 55 ⇒ 2x = 55 - 5 ⇒ 2x = 50 ⇒ x = 50/2  ⇒ x = 25  Therefore, the multiples of 5, i.e., x + 5 = 25 + 5 = 30 Therefore, the two consecutive multiples of 5 whose sum is 55 are 25 and 30.  

8. The difference in the measures of two complementary angles is 12°. Find the measure of the angles.  Solution: Let the angle be x.  Complement of x = 90 - x Given their difference = 12° Therefore, (90 - x) - x = 12° ⇒ 90 - 2x = 12 ⇒ -2x = 12 - 90 ⇒ -2x = -78 ⇒ 2x/2 = 78/2 ⇒ x = 39 Therefore, 90 - x = 90 - 39 = 51  Therefore, the two complementary angles are 39° and 51°

9. The cost of two tables and three chairs is $705. If the table costs $40 more than the chair, find the cost of the table and the chair.  Solution: The table cost $ 40 more than the chair.  Let us assume the cost of the chair to be x.  Then the cost of the table = $ 40 + x The cost of 3 chairs = 3 × x = 3x and the cost of 2 tables 2(40 + x)  Total cost of 2 tables and 3 chairs = $705 Therefore, 2(40 + x) + 3x = 705 80 + 2x + 3x = 705 80 + 5x = 705 5x = 705 - 80 5x = 625/5 x = 125 and 40 + x = 40 + 125 = 165 Therefore, the cost of each chair is $125 and that of each table is $165. 

10. If 3/5 ᵗʰ of a number is 4 more than 1/2 the number, then what is the number?  Solution: Let the number be x, then 3/5 ᵗʰ of the number = 3x/5 Also, 1/2 of the number = x/2  According to the question,  3/5 ᵗʰ of the number is 4 more than 1/2 of the number.  ⇒ 3x/5 - x/2 = 4 ⇒ (6x - 5x)/10 = 4 ⇒ x/10 = 4 ⇒ x = 40 The required number is 40.  

Try to follow the methods of solving word problems on linear equations and then observe the detailed instruction on the application of equations to solve the problems.

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Word Problems Linear Equations

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\(\textbf{1)}\) Joe and Steve are saving money. Joe starts with $105 and saves $5 per week. Steve starts with $5 and saves $15 per week. After how many weeks do they have the same amount of money? Show Equations \(y= 5x+105,\,\,\,y=15x+5\) Show Answer 10 weeks ($155)

\(\textbf{2)}\) mike and sarah collect rocks. together they collected 50 rocks. mike collected 10 more rocks than sarah. how many rocks did each of them collect show equations \(m+s=50,\,\,\,m=s+10\) show answer mike collected 30 rocks, sarah collected 20 rocks., \(\textbf{3)}\) in a classroom the ratio of boys to girls is 2:3. there are 25 students in the class. how many are girls show equations \(b+g=50,\,\,\,3b=2g\) show answer 15 girls (10 boys), \(\textbf{4)}\) kyle makes sandals at home. the sandal making tools cost $100 and he spends $10 on materials for each sandal. he sells each sandal for $30. how many sandals does he have to sell to break even show equations \(c=10x+100,\,\,\,r=30x\) show answer 5 sandals ($150), \(\textbf{5)}\) molly is throwing a beach party. she still needs to buy beach towels and beach balls. towels are $3 each and beachballs are $4 each. she bought 10 items in total and it cost $34. how many beach balls did she get show equations show answer 4 beachballs (6 towels), \(\textbf{6)}\) anna volunteers at a pet shelter. they have cats and dogs. there are 36 pets in total at the shelter, and the ratio of dogs to cats is 4:5. how many cats are at the shelter show equations \(c+d=40,\,\,\,5d=4c\) show answer 20 cats (16 dogs), \(\textbf{7)}\) a store sells oranges and apples. oranges cost $1.00 each and apples cost $2.00 each. in the first sale of the day, 15 fruits were sold in total, and the price was $25. how many of each type of frust was sold show equations \(o+a=15,\,\,\,1o+2a=25\) show answer 10 apples and 5 oranges, \(\textbf{8)}\) the ratio of red marbles to green marbles is 2:7. there are 36 marbles in total. how many are red show equations \(r+g=36,\,\,\,7r=2g\) show answer 8 red marbles (28 green marbles), \(\textbf{9)}\) a tennis club charges $100 to join the club and $10 for every hour using the courts. write an equation to express the cost \(c\) in terms of \(h\) hours playing tennis. show equation the equation is \(c=10h+100\), \(\textbf{10)}\) emma and liam are saving money. emma starts with $80 and saves $10 per week. liam starts with $120 and saves $6 per week. after how many weeks will they have the same amount of money show equations \(e = 10x + 80,\,\,\,l = 6x + 120\) show answer 10 weeks ($180 each), \(\textbf{11)}\) mark and lisa collect stamps. together they collected 200 stamps. mark collected 40 more stamps than lisa. how many stamps did each of them collect show equations \(m + l = 200,\,\,\,m = l + 40\) show answer mark collected 120 stamps, lisa collected 80 stamps., \(\textbf{12)}\) in a classroom, the ratio of boys to girls is 3:5. there are 40 students in the class. how many are boys show equations \(b + g = 40,\,\,\,5b = 3g\) show answer 15 boys (25 girls), \(\textbf{13)}\) lisa is selling handmade jewelry. the materials cost $60, and she sells each piece for $20. how many pieces does she have to sell to break even show equations \(c=60,\,\,\,r=20x\) show answer 3 pieces, \(\textbf{14)}\) tom is buying books and notebooks for school. books cost $15 each, and notebooks cost $3 each. he bought 12 items in total, and it cost $120. how many notebooks did he buy show equations \(b + n = 12,\,\,\,15b+3n=120\) show answer 5 notebooks (7 books), \(\textbf{15)}\) emily volunteers at an animal shelter. they have rabbits and guinea pigs. there are 36 animals in total at the shelter, and the ratio of guinea pigs to rabbits is 4:5. how many guinea pigs are at the shelter show equations \(r + g = 36,\,\,\,5g=4r\) show answer 16 guinea pigs (20 rabbits), \(\textbf{16)}\) mike and sarah are going to a theme park. mike’s ticket costs $40, and sarah’s ticket costs $30. they also bought $20 worth of food. how much did they spend in total show equations \(m + s + f = t,\,\,\,m=40,\,\,\,s=30,\,\,\,f=20\) show answer they spent $90 in total., \(\textbf{17)}\) the ratio of red marbles to blue marbles is 2:3. there are 50 marbles in total. how many are blue show equations \(r + b = 50,\,\,\,3r=2b\) show answer 30 blue marbles (20 red marbles), \(\textbf{18)}\) a pizza restaurant charges $12 for a large pizza and $8 for a small pizza. if a customer buys 5 pizzas in total, and it costs $52, how many large pizzas did they buy show equations \(l + s = 5,\,\,\,12l+8s=52\) show answer they bought 3 large pizzas (2 small pizzas)., \(\textbf{19)}\) the area of a rectangle is 48 square meters. if the length is 8 meters, what is the width of the rectangle show equations \(a=l\times w,\,\,\,l=8,\,\,\,a=48\) show answer the width is 6 meters., \(\textbf{20)}\) two numbers have a sum of 50. one number is 10 more than the other. what are the two numbers show equations \(x+y=50,\,\,\,x=y+10\) show answer the numbers are 30 and 20., \(\textbf{21)}\) a store sells jeans for $40 each and t-shirts for $20 each. in the first sale of the day, they sold 8 items in total, and the price was $260. how many of each type of item was sold show equations \(j+t=8,\,\,\,40j+20t=260\) show answer 5 jeans and 3 t-shirts were sold., \(\textbf{22)}\) the ratio of apples to carrots is 3:4. there are 28 fruits in total. how many are apples show equations \(\)a+c=28,\,\,\,4a=3c show answer there are 12 apples and 16 carrots., \(\textbf{23)}\) a phone plan costs $30 per month, and there is an additional charge of $0.10 per minute for calls. write an equation to express the cost \(c\) in terms of \(m\) minutes. show equation the equation is \(\)c=30+0.10m, \(\textbf{24)}\) a triangle has a base of 8 inches and a height of 6 inches. calculate its area. show equations \(a=0.5\times b\times h,\,\,\,b=8,\,\,\,h=6\) show answer the area is 24 square inches., \(\textbf{25)}\) a store sells shirts for $25 each and pants for $45 each. in the first sale of the day, 4 items were sold, and the price was $180. how many of each type of item was sold show equations \(t+p=4,\,\,\,25t+45p=180\) show answer 0 shirts and 4 pants were sold., \(\textbf{26)}\) a garden has a length of 12 feet and a width of 10 feet. calculate its area. show equations \(a=l\times w,\,\,\,l=12,\,\,\,w=10\) show answer the area is 120 square feet., \(\textbf{27)}\) the sum of two consecutive odd numbers is 56. what are the two numbers show equations \(x+y=56,\,\,\,x=y+2\) show answer the numbers are 27 and 29., \(\textbf{28)}\) a toy store sells action figures for $15 each and toy cars for $5 each. in the first sale of the day, 10 items were sold, and the price was $110. how many of each type of item was sold show equations \(a+c=10,\,\,\,15a+5c=110\) show answer 6 action figures and 4 toy cars were sold., \(\textbf{29)}\) a bakery sells pie for $2 each and cookies for $1 each. in the first sale of the day, 14 items were sold, and the price was $25. how many of each type of item was sold show equations \(p+c=14,\,\,\,2p+c=25\) show answer 11 pies and 3 cookies were sold., \(\textbf{for 30-33}\) two car rental companies charge the following values for x miles. car rental a: \(y=3x+150 \,\,\) car rental b: \(y=4x+100\), \(\textbf{30)}\) which rental company has a higher initial fee show answer company a has a higher initial fee, \(\textbf{31)}\) which rental company has a higher mileage fee show answer company b has a higher mileage fee, \(\textbf{32)}\) for how many driven miles is the cost of the two companies the same show answer the companies cost the same if you drive 50 miles., \(\textbf{33)}\) what does the \(3\) mean in the equation for company a show answer for company a, the cost increases by $3 per mile driven., \(\textbf{34)}\) what does the \(100\) mean in the equation for company b show answer for company b, the initial cost (0 miles driven) is $100., \(\textbf{for 35-39}\) andy is going to go for a drive. the formula below tells how many gallons of gas he has in his car after m miles. \(g=12-\frac{m}{18}\), \(\textbf{35)}\) what does the \(12\) in the equation represent show answer andy has \(12\) gallons in his car when he starts his drive., \(\textbf{36)}\) what does the \(18\) in the equation represent show answer it takes \(18\) miles to use up \(1\) gallon of gas., \(\textbf{37)}\) how many miles until he runs out of gas show answer the answer is \(216\) miles, \(\textbf{38)}\) how many gallons of gas does he have after 90 miles show answer the answer is \(7\) gallons, \(\textbf{39)}\) when he has \(3\) gallons remaining, how far has he driven show answer the answer is \(162\) miles, \(\textbf{for 40-42}\) joe sells paintings. each month he makes no commission on the first $5,000 he sells but then makes a 10% commission on the rest., \(\textbf{40)}\) find the equation of how much money x joe needs to sell to earn y dollars per month. show answer the answer is \(y=.1(x-5,000)\), \(\textbf{41)}\) how much does joe need to sell to earn $10,000 in a month. show answer the answer is \($105,000\), \(\textbf{42)}\) how much does joe earn if he sells $45,000 in a month show answer the answer is \($4,000\), see related pages\(\), \(\bullet\text{ word problems- linear equations}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- averages}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- consecutive integers}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- distance, rate and time}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- break even}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- ratios}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- age}\) \(\,\,\,\,\,\,\,\,\), \(\bullet\text{ word problems- mixtures and concentration}\) \(\,\,\,\,\,\,\,\,\), linear equations are a type of equation that has a linear relationship between two variables, and they can often be used to solve word problems. in order to solve a word problem involving a linear equation, you will need to identify the variables in the problem and determine the relationship between them. this usually involves setting up an equation (or equations) using the given information and then solving for the unknown variables . linear equations are commonly used in real-life situations to model and analyze relationships between different quantities. for example, you might use a linear equation to model the relationship between the cost of a product and the number of units sold, or the relationship between the distance traveled and the time it takes to travel that distance. linear equations are typically covered in a high school algebra class. these types of problems can be challenging for students who are new to algebra, but they are an important foundation for more advanced math concepts. one common mistake that students make when solving word problems involving linear equations is failing to set up the problem correctly. it’s important to carefully read the problem and identify all of the relevant information, as well as any given equations or formulas that you might need to use. other related topics involving linear equations include graphing and solving systems. understanding linear equations is also useful for applications in fields such as economics, engineering, and physics., about andymath.com, andymath.com is a free math website with the mission of helping students, teachers and tutors find helpful notes, useful sample problems with answers including step by step solutions, and other related materials to supplement classroom learning. if you have any requests for additional content, please contact andy at [email protected] . he will promptly add the content. topics cover elementary math , middle school , algebra , geometry , algebra 2/pre-calculus/trig , calculus and probability/statistics . in the future, i hope to add physics and linear algebra content. visit me on youtube , tiktok , instagram and facebook . andymath content has a unique approach to presenting mathematics. the clear explanations, strong visuals mixed with dry humor regularly get millions of views. we are open to collaborations of all types, please contact andy at [email protected] for all enquiries. to offer financial support, visit my patreon page. let’s help students understand the math way of thinking thank you for visiting. how exciting.

Solving Linear Equations

Solving linear equations means finding the value of the variable(s) given in the linear equations. A linear equation is a combination of an algebraic expression and an equal to (=) symbol. It has a degree of 1 or it can be called a first-degree equation. For example, x + y = 4 is a linear equation. Sometimes, we may have to find the values of variables involved in a linear equation. When we are given two or more such linear equations, we can find the values of each variable by solving linear equations. There are a few methods to solve linear equations. Let us discuss each of these methods in detail.

Solving Linear Equations in One Variable

A linear equation in one variable is an equation of degree one and has only one variable term. It is of the form 'ax+b = 0', where 'a' is a non zero number and 'x' is a variable. By solving linear equations in one variable, we get only one solution for the given variable. An example for this is 3x - 6 = 0. The variable 'x' has only one solution, which is calculated as 3x - 6 = 0 3x = 6 x = 6/3 x = 2

For solving linear equations with one variable, simplify the equation such that all the variable terms are brought to one side and the constant value is brought to the other side. If there are any fractional terms then find the LCM ( Least Common Multiple ) and simplify them such that the variable terms are on one side and the constant terms are on the other side. Let us work out a small example to understand this.

4x + 8 = 8x - 10. To find the value of 'x', let us simplify and bring the 'x' terms to one side and the constant terms to another side.

4x - 8x = -10 - 8 -4x = -18 4x = 18 x = 18/4 On simplifying, we get x = 9/2.

Solving Linear Equations by Substitution Method

The substitution method is one of the methods of solving linear equations. In the substitution method , we rearrange the equation such that one of the values is substituted in the second equation. Now that we are left with an equation that has only one variable, we can solve it and find the value of that variable. In the two given equations, any equation can be taken and the value of a variable can be found and substituted in another equation. For solving linear equations using the substitution method, follow the steps mentioned below. Let us understand this with an example of solving the following system of linear equations. x + y = 6 --------------(1) 2x + 4y = 20 -----------(2)

Step 1: Find the value of one of the variables using any one of the equations. In this case, let us find the value of 'x' from equation (1). x + y = 6 ---------(1) x = 6 - y Step 2: Substitute the value of the variable found in step 1 in the second linear equation. Now, let us substitute the value of 'x' in the second equation 2x + 4y = 20.

x = 6 - y Substituting the value of 'x' in 2x + 4y = 20, we get,

2(6 - y) + 4y = 20 12 - 2y + 4y = 20 12 + 2y = 20 2y = 20 - 12 2y = 8 y = 8/2 y = 4 Step 3: Now substitute the value of 'y' in either equation (1) or (2). Let us substitute the value of 'y' in equation (1).

x + y = 6 x + 4 = 6 x = 6 - 4 x = 2 Therefore, by substitution method, the linear equations are solved, and the value of x is 2 and y is 4.

Solving Linear Equations by Elimination Method

The elimination method is another way to solve a system of linear equations. Here we make an attempt to multiply either the 'x' variable term or the 'y' variable term with a constant value such that either the 'x' variable terms or the 'y' variable terms cancel out and gives us the value of the other variable. Let us understand the steps of solving linear equations by elimination method . Consider the given linear equations: 2x + y = 11 ----------- (1) x + 3y = 18 ---------- (2) Step 1: Check whether the terms are arranged in a way such that the 'x' term is followed by a 'y' term and an equal to sign and after the equal to sign the constant term should be present. The given set of linear equations are already arranged in the correct way which is ax+by=c or ax+by-c=0.

Step 2: The next step is to multiply either one or both the equations by a constant value such that it will make either the 'x' terms or the 'y' terms cancel out which would help us find the value of the other variable. Now in equation (2), let us multiply every term by the number 2 to make the coefficients of x the same in both the equations. x + 3y = 18 ---------- (2) Multiplying all the terms in equation (2) by 2, we get,

2(x) + 2(3y) = 2(18). Now equation (2) becomes, 2x + 6y = 36 -----------(2)

Therefore, y = 5. Step 4: Using the value obtained in step 3, find out the value of another variable by substituting the value in any of the equations. Let us substitute the value of 'y' in equation (1). We get, 2x + y = 11 2x + 5 = 11 2x = 11 - 5 2x = 6 x = 6/2 x = 3

Therefore, by solving linear equations, we get the value of x = 3 and y = 5.

Graphical Method of Solving Linear Equations

Another method for solving linear equations is by using the graph. When we are given a system of linear equations, we graph both the equations by finding values for 'y' for different values of 'x' in the coordinate system. Once it is done, we find the point of intersection of these two lines. The (x,y) values at the point of intersection give the solution for these linear equations. Let us take two linear equations and solve them using the graphical method.

x + y = 8 -------(1)

y = x + 2 --------(2)

Let us take some values for 'x' and find the values for 'y' for the equation x + y = 8. This can also be rewritten as y = 8 - x.

Let us take some values for 'x' and find the values for 'y' in the equation y = x + 2.

Plotting these points on the coordinate plane, we get a graph like this.

Now, we find the point of intersection of these lines to find the values of 'x' and 'y'. The two lines intersect at the point (3,5). Therefore, x = 3 and y = 5 by using the graphical method of solving linear equations .

This method is also used to find the optimal solution of linear programming problems. Let us look at one more method of solving linear equations, which is the cross multiplication method.

Cross Multiplication Method of Solving Linear Equations

The cross multiplication method enables us to solve linear equations by picking the coefficients of all the terms ('x' , 'y' and the constant terms) in the format shown below and apply the formula for finding the values of 'x' and 'y'.

Topics Related to Solving Linear Equations

Check the given articles related to solving linear equations.

• Linear Equations
• Application of Linear Equations
• Two-Variable Linear Equations
• Linear Equations and Half Planes
• One Variable Linear Equations and Inequations

Solving Linear Equations Examples

Example 1: Solve the following linear equations by the substitution method.

3x + y = 13 --------- (1) 2x + 3y = 18 -------- (2)

By using the substitution method of solving linear equations, let us take the first equation and find the value of 'y' and substitute it in the second equation.

From equation (1), y = 13-3x. Now, substituting the value of 'y' in equation (2), we get, 2x + 3 (13 - 3x) = 18 2x + 39 - 9x = 18 -7x + 39 = 18 -7x = 18 - 39 -7x = -21 x = -21/-7 x = 3 Now, let us substitute the value of 'x = 3' in equation (1) and find the value of 'y'. 3x + y = 13 ------- (1) 3(3) + y = 13 9 + y = 13 y = 13 - 9 y = 4

Therefore, by the substitution method, the value of x is 3 and y is 4.

Example 2: Using the elimination method of solving linear equations find the values of 'x' and 'y'.

3x + y = 21 ------ (1) 2x + 3y = 28 -------- (2)

By using the elimination method, let us make the 'y' variable to be the same in both the equations (1) and (2). To do this let us multiply all the terms of the first equation by 3. Therefore equation (1) becomes,

3(3x) + 3(y) = 63 9x + 3y = 63 ---------- (3) The second equation is, 2x + 3y = 28 Now let us cancel the 'y' terms and find the value of 'x' by subtracting equation (2) from equation (3). This is done by changing the signs of all the terms in equation (2).

Example 3: Using the cross multiplication method of solving linear equations, solve the following equations.

x + 2y - 16 = 0 --------- (1) 4x - y - 10 = 0 ---------- (2)

Compare the given equation with \(a_{1}\)x + \(b_{1}\)y + \(c_{1}\) = 0, and \(a_{2}\)x+\(b_{2}\)y+\(c_{2}\) = 0. From the given equations,

\(a_{1}\) = 1, \(a_{2}\) = 4, \(b_{1}\) = 2, \(b_{2}\) = -1, \(c_{1}\) = -16, and \(c_{2}\) = -10.

By cross multiplication method,

x = \(b_{1}\)\(c_{2}\) - \(b_{2}\)\(c_{1}\)/\(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\) y = \(c_{1}\)\(a_{2}\) - \(c_{2}\)\(a_{1}\) / \(a_{1}\)\(b_{2}\) - \(a_{2}\)\(b_{1}\)

Substituting the values in the formula we get,

x = ((2)(-10)) - ((-1)(-16)) / ((1)(-1)) - ((4)(2)) x = (-20-16)/(-1-8) x = -36/-9 x = 36/9 x = 4 y = ((-16)(4)) - ((-10)(1)) / ((1)(-1)) - ((4)(2)) y = (-64 + 10) / (-1 - 8) y = -54 / -9 y = 54/9 y = 6 Therefore, by the cross multiplication method, the value of x is 4 and y is 6.

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Practice Questions on Solving Linear Equations

Faqs on solving linear equations, what does it mean by solving linear equations.

An equation that has a degree of 1 is called a linear equation. We can have one variable linear equations , two-variable linear equations , linear equations with three variables, and more depending on the number of variables in it. Solving linear equations means finding the values of all the variables present in the equation. This can be done by substitution method, elimination method, graphical method, and the cross multiplication method . All these methods are different ways of finding the values of the variables.

How to Use the Substitution Method for Solving Linear Equations?

The substitution method of solving equations states that for a given system of linear equations, find the value of either 'x' or 'y' from any of the given equations and then substitute the value found of 'x' or 'y' in another equation so that the other unknown value can be found.

How to Use the Elimination Method for Solving Linear Equations?

In the elimination method of solving linear equations, we multiply a constant or a number with one equation or both the equations such that either the 'x' terms or the 'y' terms are the same. Then we cancel out the same term in both the equations by either adding or subtracting them and find the value of one variable (either 'x' or 'y'). After finding one of the values, we substitute the value in one of the equations and find the other unknown value.

What is the Graphical Method of Solving Linear Equations?

In the graphical method of solving linear equations, we find the value of 'y' from the given equations by putting the values of x as 0, 1, 2, 3, and so on, and plot a graph in the coordinate system for the line for various values of 'x' for both the system of linear equations. We will see that these two lines intersect at a point. This point is the solution for the given system of linear equations. If there is no intersection point between two lines, then we consider them as parallel lines , and if we found that both the lines lie on each other, those are known as coincident lines and have infinitely many solutions.

What are the Steps of Solving Linear Equations that has One Variable?

A linear equation is an equation with degree 1. To solve a linear equation that has one variable we bring the variable to one side and the constant value to the other side. Then, a non-zero number may be added, subtracted, multiplied, or divided on both sides of the equation. For example, a linear equation with one variable will be of the form 'x - 4 = 2'. To find the value of 'x', we add the constant value '4' to both sides of the equation. Therefore, the value of 'x = 6'.

What are the Steps of Solving Linear Equations having Three Variables?

To solve a system of linear equations that has three variables, we take any two equations and variables. We then take another pair of linear equations and also solve for the same variable. Now that, we have two linear equations with two variables, we can use the substitution method or elimination method, or any other method to solve the values of two unknown variables. After finding these two variables, we substitute them in any of the three equations to find the third unknown variable.

What are the 4 Methods of Solving Linear Equations?

The methods for solving linear equations are given below:

• Substitution method
• Elimination method
• Cross multiplication method
• Graphical method

5.2 Linear Equations in One Variable with Applications

Learning objectives.

After completing this section, you should be able to:

• Solve linear equations in one variable using properties of equations.
• Construct a linear equation to solve applications.
• Determine equations with no solution or infinitely many solutions.
• Solve a formula for a given variable.

In this section, we will study linear equations in one variable. There are several real-world scenarios that can be represented by linear equations: taxi rentals with a flat fee and a rate per mile; cell phone bills that charge a monthly fee plus a separate rate per text; gym memberships with a monthly fee plus a rate per class taken; etc. For example, if you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month?

Linear Equations and Applications

Solving any equation is like discovering the answer to a puzzle. The purpose of solving an equation is to find the value or values of the variable that makes the equation a true statement. Any value of the variable that makes the equation true is called a solution to the equation. It is the answer to the puzzle! There are many types of equations that we will learn to solve. In this section, we will focus on a linear equation , which is an equation in one variable that can be written as

where a a and b b are real numbers and a ≠ 0 a ≠ 0 , such that a a is the coefficient of x x and b b is the constant.

To solve a linear equation, it is a good idea to have an overall strategy that can be used to solve any linear equation. In the Example 5.12, we will give the steps of a general strategy for solving any linear equation. Simplifying each side of the equation as much as possible first makes the rest of the steps easier.

Example 5.12

Solving a linear equation using a general strategy.

Solve 7 ( n − 3 ) − 8 = − 15 7 ( n − 3 ) − 8 = − 15

Your Turn 5.12

In Example 5.12 , we used both the addition and division property of equations. All the properties of equations are summarized in table below. Basically, what you do to one side of the equation, you must do to the other side of the equation to preserve equality.

Be careful to multiply and divide every term on each side of the equation. For example, 2 + x = x 3 2 + x = x 3 is solved by multiplying BOTH sides of the equation by 3 to get 3 ( 2 + x ) = 3 ( x 3 ) 3 ( 2 + x ) = 3 ( x 3 ) which gives 6 + 3 x = x 6 + 3 x = x . Using parentheses will help you remember to use the distributive property! A division example, such as 3 ( x + 2 ) = 6 x + 9 3 ( x + 2 ) = 6 x + 9 , can be solved by dividing BOTH sides of the equation by 3 to get 3 ( x + 2 ) 3 = 6 x + 9 3 , 3 ( x + 2 ) 3 = 6 x + 9 3 , which then will lead to x + 2 = 2 x + 3 x + 2 = 2 x + 3 .

Example 5.13

Solving a linear equation using properties of equations.

Solve 9 ( y − 2 ) − y = 16 + 7 y 9 ( y − 2 ) − y = 16 + 7 y .

Step 1: Simplify each side.

Step 2: Collect all variables on one side.

Step 3: Collect constant terms on one side.

Step 4: Make the coefficient of the variable 1. Already done!

Step 5: Check.

Your Turn 5.13

Who invented the symbol for equals .

Before the creation of a symbol for equality, it was usually expressed with a word that meant equals, such as aequales (Latin), esgale (French), or gleich (German). Welsh mathematician and physician Robert Recorde is given credit for inventing the modern sign. It first appears in writing in The Whetstone of Witte , a book Recorde wrote about algebra, which was published in 1557. In this book, Recorde states, "I will set as I do often in work use, a pair of parallels, or Gemowe (twin) lines of one length, thus: ===, because no two things can be more equal." Although his version of the sign was a bit longer than the one we use today, his idea stuck and "=" is used throughout the world to indicate equality in mathematics.

In Algebraic Expressions , you translated an English sentence into an equation. In this section, we take that one step further and translate an English paragraph into an equation, and then we solve the equation. We can go back to the opening question in this section: If you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month? We can create an equation for this scenario and then solve the equation (see Example 5.15 ).

Example 5.14

Constructing a linear equation to solve an application.

The Beaudrie family has two cats, Basil and Max. Together, they weigh 23 pounds. Basil weighs 16 pounds. How much does Max weigh?

Let b b = Basil’s weight and m m = Max’s weight.

We also know that Basil weighs 16 pounds so:

Steps 1 and 2: 16 + m = 23 16 + m = 23

Since both sides are simplified, the variable is on one side of the equation, we start in Step 3 and collect the constants on one side:

Step 4: is already done so we go to Step 5:

Basil weighs 16 pounds and Max weighs 7 pounds.

Your Turn 5.14

Example 5.15, constructing a linear equation to solve another application.

If you join your local gym at $10 per month and pay $5 per class, how many classes can you take if your gym budget is $75 per month?

If we let x x = number of classes, the expression 5 x + 10 5 x + 10 would represent what you pay per month if each class is $5 and there’s a $10 monthly fee per class. $10 is your constant. If you want to know how many classes you can take if you have a $75 monthly gym budget, set the equation equal to 75. Then solve the equation 5 x + 10 = 75 5 x + 10 = 75 for x x .

Steps 1 and 2:

The solution is 13 classes. You can take 13 classes on a $75 monthly gym budget.

Your Turn 5.15

Example 5.16, constructing an application from a linear equation.

Write an application that can be solved using the equation 50 x + 35 = 185 50 x + 35 = 185 . Then solve your application.

Answers will vary. Let’s say you want to rent a snowblower for a huge winter storm coming up. If x x = the number of days you rent a snowblower, then the expression 50 x + 35 50 x + 35 represents what you pay if, for each day, it costs $50 to rent the snowblower and there is a $35 flat rental fee. $35 is the constant. To find out how many days you can rent a snowblower for $185, set the expression equal to 185. Then solve the equation 50 x + 35 = 185 50 x + 35 = 185 for x x .

The equation is 50 x + 35 = 185 50 x + 35 = 185 and the solution is 3 days. You can rent a snowblower for 3 days on a $185 budget.

Your Turn 5.16

Linear equations with no solutions or infinitely many solutions.

Every linear equation we have solved thus far has given us one numerical solution. Now we'll look at linear equations for which there are no solutions or infinitely many solutions.

Example 5.17

Solving a linear equation with no solution.

Solve 3 ( x + 4 ) = 4 x + 8 − x 3 ( x + 4 ) = 4 x + 8 − x .

Step 1: Simplify each side. 3 ( x + 4 ) = 4 x + 8 − x 3 ( x + 4 ) = 4 x + 8 − x

Step 2: Collect all variables to one side. 3 x + 12 − 3 x = 3 x + 8 − 3 x 3 x + 12 − 3 x = 3 x + 8 − 3 x

The variable x x disappeared! When this happens, you need to examine what remains. In this particular case, we have 12 = 8 12 = 8 , which is not a true statement. When you have a false statement, then you know the equation has no solution; there does not exist a value for x x that can be put into the equation that will make it true.

Your Turn 5.17

Example 5.18, solving a linear equation with infinitely many solutions.

Solve 2 ( x + 5 ) = 4 ( x + 3 ) − 2 x − 2 2 ( x + 5 ) = 4 ( x + 3 ) − 2 x − 2 .

As with the previous example, the variable disappeared. In this case, however, we have a true statement ( 10 = 10 10 = 10 ). When this occurs we say there are infinitely many solutions; any value for x x will make this statement true.

Your Turn 5.18

Solving a formula for a given variable.

You are probably familiar with some geometry formulas. A formula is a mathematical description of the relationship between variables. Formulas are also used in the sciences, such as chemistry, physics, and biology. In medicine they are used for calculations for dispensing medicine or determining body mass index. Spreadsheet programs rely on formulas to make calculations. It is important to be able to manipulate formulas and solve for specific variables.

To solve a formula for a specific variable means to isolate that variable on one side of the equal sign with a coefficient of 1. All other variables and constants are on the other side of the equal sign. To see how to solve a formula for a specific variable, we will start with the distance, rate, and time formula.

Example 5.19

Solving for a given variable with distance, rate, and time.

Solve the formula d = r t d = r t for t t . This is the distance formula where d d = distance, r r = rate, and t t = time.

Divide both sides by r r : d / r = r t / r d / r = r t / r

d / r = t d / r = t

Your Turn 5.19

Solving for a Variable in an Equation

Example 5.20

Solving for a given variable in the area formula for a triangle.

Solve the formula A = ½ A = ½ bh bh for h h . This is the area formula of a triangle where A A = area, b b = base, and h h = height.

Step 1: Multiply both sides by 2.

Step 2: Divide both sides by b b .

Your Turn 5.20

Work it out, using algebra to understand card tricks.

You will need to perform this card trick with another person. Before you begin, the two people must first decide which of the two will be the Dealer and which will be the Partner , as each will do something different. Once you have decided upon that, follow the steps here:

Step 1: Dealer and Partner: Take a regular deck of 52 cards, and remove the face cards and the 10s.

Step 2: Dealer and Partner: Shuffle the remaining cards

Step 3: Dealer and Partner: Select one card each, but keep them face down and don’t look at them yet.

Step 4: Dealer: Look at your card (just the Dealer!). Multiply its value by 2 (Aces = 1).

Step 5: Dealer: Add 2 to this result.

Step 6: Dealer: Multiply your answer by 5.

Step 7: Partner: Look at your card.

Step 8: Partner: Calculate: 10 - your card, and tell this information to the dealer.

Step 9: Dealer: Subtract the value the Partner tells you from your total to get a final answer.

Step 10: Dealer: verbally state the final answer.

Step 11: Dealer and Partner: Turn over your cards. Now, answer the following questions

• Did the trick work? How do you know?
• Why did this occur? In other words, how does this trick work?

Check Your Understanding

• F = 5 9 C − 32
• F = 9 5 C − 32
• F = 5 9 C + 32
• F = 9 5 C + 32
• C = K + 273
• K = C + 273
• K = C − 273
• C = K − 273
• K = 5 9 ( F − 32 ) + 273
• K = 5 9 F + 241
• K = 9 5 ( F − 32 ) + 273
• K = 9 5 F + 241
• R = 5 9 C − 492
• R = 9 5 C + 492
• R = C + 492
• R = 5 9 ( C − 492 )

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3.3: Applications of Linear Systems with Two Variables

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Learning Objectives

• Set up and solve applications involving relationships between two variables.
• Set up and solve mixture problems.
• Set up and solve uniform motion problems (distance problems).

Problems Involving Relationships between Two Variables

If we translate an application to a mathematical setup using two variables, then we need to form a linear system with two equations. Setting up word problems with two variables often simplifies the entire process, particularly when the relationships between the variables are not so clear.

Example \(\PageIndex{1}\):

The sum of \(4\) times a larger integer and \(5\) times a smaller integer is \(7\). When twice the smaller integer is subtracted from \(3\) times the larger, the result is \(11\). Find the integers.

Begin by assigning variables to the larger and smaller integer.

Let \(x\) represent the larger integer.

Let \(y\) represent the smaller integer.

When using two variables, we need to set up two equations. The first sentence describes a sum and the second sentence describes a difference.

This leads to the following system:

\(\left\{ \begin{array} { l } { 4 x + 5 y = 7 } \\ { 3 x - 2 y = 11 } \end{array} \right.\)

Solve using the elimination method. To eliminate the variable \(y\) multiply the first equation by \(2\) and the second by \(5\).

\(\left\{ \begin{array} { l l } { 4 x + 5y = 7 } & { \stackrel { \times2 } { \Rightarrow } } \\ { 3 x -2y = 11 } & { \stackrel { \Rightarrow } { \times 5 } } \end{array} \right. \left\{ \begin{array} { l } { 8 x + 10 y = 14 } \\ { 15 x -10y = 55 } \end{array} \right.\)

Add the equations in the equivalent system and solve for \(x\).

\(\begin{aligned} 8 x \color{red}{+ 10 y} &\color{black}{=} 14 \\ \pm 15 x \color{red}{- 10 y} & \color{black}{=} 55 \\ \hline\\ 23x & = 99\\ x & = \frac{69}{23}\\x&=3 \end{aligned}\)

Back substitute to find \(y\).

\(\begin{aligned} 4 x + 5 y & = 7 \\ 4 ( \color{OliveGreen}{3} \color{black}{)} + 5 y & = 7 \\ 12 + 5 y & = 7 \\ 5 y & = - 5 \\ y & = - 1 \end{aligned}\)

The largest integer is \(3\) and the smaller integer is \(-1\).

Exercise \(\PageIndex{1}\)

An integer is \(1\) less than twice that of another. If their sum is \(20\), find the integers.

The two integers are \(7\) and \(13\).

www.youtube.com/v/LnzO1_J4X20

Next consider applications involving simple interest and money.

A total of \($12,800\) was invested in two accounts. Part was invested in a CD at a \(3 \frac{1}{8}\)% annual interest rate and part was invested in a money market fund at a \(4 \frac{3}{4}\)% annual interest rate. If the total simple interest for one year was \($465\), then how much was invested in each account?

Begin by identifying two variables.

Let \(x\) represent the amount invested at \(3 \frac{1}{8}\)% \(= 3.125\) % \(= 0.03125\).

Let \(y\) represent the amount invested at \(4 \frac{3}{4}\)% \(= 4.75\)% = \(0.0475\).

The total amount in both accounts can be expressed as

\(x+y=12,800\)

To set up a second equation, use the fact that the total interest was \($465\). Recall that the interest for one year is the interest rate times the principal \((I = prt = pr ⋅ 1 = p)\). Use this to add the interest in both accounts. Be sure r to use the decimal equivalents for the interest rates given as percentages.

\(\begin{aligned} \color{Cerulean} { interest\: from\: the\: C D\: +\: interest\: from\: the\: fund\: =\: total\: interest } \\ 0.03125 x \quad\quad\quad +\quad\quad\:\: \quad 0.0475 y \quad\quad\quad\quad = 465\quad\quad\quad\quad\:\: \end{aligned}\)

These two equations together form the following linear system:

\(\left\{ \begin{array} { c } { x + y = 12,800 } \\ { 0.03125 x + 0.0475 y = 465 } \end{array} \right.\)

Eliminate \(x\) by multiplying the first equation by \(-0.03125\).

Next, add the resulting equations.

\(\begin{aligned} \color{red}{- 0.03125 x}\color{black}{ -} 0.03125 y &= - 400 \\ \pm\:\: \color{red}{0.03125 x}\color{black}{ +} 0.0475 y &= 465 \\ \hline \\0.01625y &=65 \\ y & = \frac{65}{0.01635} \\ y & = 4,000 \end{aligned}\)

Back substitute to find \(x\).

\(\begin{aligned} x + y & = 12,800 \\ x + 4000 & = 12,800 \\ x & = 8,800 \end{aligned}\)

\($4,000\) was invested at \(4 \frac{3}{4}\)% and \($8,800\) was invested at \(3 \frac{1}{8}\)%.

Example \(\PageIndex{3}\):

A jar consisting of only nickels and dimes contains \(58\) coins. If the total value is \($4.20\), how many of each coin is in the jar?

Let \(n\) represent the number of nickels in the jar.

Let \(d\) represent the number of dimes in the jar.

The total number of coins in the jar can be expressed using the following equation:

Next, use the value of each coin to determine the total value \($4.20\).

\(\begin{aligned} \color{Cerulean} {value\: of\: nickels\: + \: value\: of\: dimes\: =\: total\:value} \\ 0.05 n\quad\quad\: +\quad\:\:\: 0.10 d\quad\quad\quad = 4.20\quad\quad\quad \end{aligned}\)

This leads us to the following linear system:

\(\left\{ \begin{array} { l } { n + d = 58 } \\ { 0.05 n + 0.10 d = 4.20 } \end{array} \right.\)

Here we will solve using the substitution method. In the first equation, we can solve for \(n\).

Substitute \(n = 58 − d\) into the second equation and solve for \(d\).

\(\begin{aligned} 0.05 ( \color{Cerulean}{58 - d}\color{black}{ )} + 0.10 d & = 4.20 \\ 2.9 - 0.05 d + 0.10 d & = 4.20 \\ 2.9 + 0.05 d & = 4.20 \\ 0.05 d & = 1.3 \\ d & = 26 \end{aligned}\)

Now back substitute to find the number of nickels.

\(\begin{aligned} n & = 58 - d \\ & = 58 - 26 \\ & = 32 \end{aligned}\)

There are \(32\) nickels and \(26\) dimes in the jar.

Exercise \(\PageIndex{2}\)

Joey has a jar full of \(40\) coins consisting of only quarters and nickels. If the total value is \($5.00\), how many of each coin does Joey have?

Joey has \(15\) quarters and \(25\) nickels.

www.youtube.com/v/41bxt_tThkA

Mixture Problems

Mixture problems often include a percentage and some total amount. It is important to make a distinction between these two types of quantities. For example, if a problem states that a \(20\)-ounce container is filled with a \(2\)% saline (salt) solution, then this means that the container is filled with a mixture of salt and water as follows:

In other words, we multiply the percentage times the total to get the amount of each part of the mixture.

Example \(\PageIndex{4}\):

A \(1.8\)% saline solution is to be combined and mixed with a \(3.2\)% saline solution to produce \(35\) ounces of a \(2.2\)% saline solution. How much of each is needed?

Let \(x\) represent the amount of \(1.8\)% saline solution needed.

Let \(y\) represent the amount of \(3.2\)% saline solution needed.

The total amount of saline solution needed is \(35\) ounces. This leads to one equation,

The second equation adds up the amount of salt in the correct percentages. The amount of salt is obtained by multiplying the percentage times the amount, where the variables \(x\) and \(y\) represent the amounts of the solutions. The amount of salt in the end solution is \(2.2\)% of the \(35\) ounces, or \(.022(35)\).

\(\begin{aligned} \color {Cerulean} { salt\: in\: 1.8} \%\: \color{Cerulean}{solution } + \color{Cerulean} { salt\: in \:} 3.2 \% \color{Cerulean} { \:solution } = \color{Cerulean} { salt\: in\: the\: end\: solution } \\ 0.018 x \quad\quad\quad+ \quad\:\quad 0.032 y\quad\quad\quad\quad = \quad\quad\quad0.022 ( 35 )\quad\quad\quad \end{aligned}\)

The algebraic setup consists of both equations presented as a system:

\(\left\{ \begin{array} { c } { x + y = 35 } \\ { 0.018 x + 0.032 y = 0.022 ( 35 ) } \end{array} \right.\)

Add the resulting equations together

\(\begin{aligned} - 0.018 x - 0.018 y &= - 0.63 \\ \pm\:\: 0.018 x + 0.032 y &= 0.77 \\ \hline \\0.014y &=0.14\\y&=\frac{0.14}{0.014}\\y&=10 \end{aligned}\)

\(\begin{aligned} x + y & = 35 \\ x + \color{OliveGreen}{10} & \color{Black}{=} 35 \\ x & = 25 \end{aligned}\)

We need \(25\) ounces of the \(1.8\)% saline solution and \(10\) ounces of the \(3.2\)% saline solution.

Example \(\PageIndex{5}\):

An \(80\)% antifreeze concentrate is to be mixed with water to produce a \(48\)-liter mixture containing \(25\)% antifreeze. How much water and antifreeze concentrate is needed?

Let \(x\) represent the amount of \(80\)% antifreeze concentrate needed.

Let \(y\) represent the amount of water needed.

The total amount of the mixture must be \(48\) liters.

The second equation adds up the amount of antifreeze from each solution in the correct percentages. The amount of antifreeze in the end result is \(25\)% of \(48\) liters, or \(0.25(48)\).

\(\begin{aligned} \color {Cerulean} { antifreeze\: in\: 80} \%\: \color{Cerulean}{concentrate } + \color{Cerulean} { antrifreeze\: in \: water} = \color{Cerulean} { antifreeze\: in\: the\: end\: mixture } \\ 0.018 x \quad\quad \quad\quad\quad+ \: \quad\:\quad 0.032 y\quad\quad\quad\quad = \quad\quad\quad\quad0.022 ( 35 )\quad\quad\quad\quad\quad \end{aligned}\)

Now we can form a system of two linear equations and two variables as follows:

\(\left\{ \begin{array} { c } { x + y = 48 } \\ { 0.80 x = 0.25 ( 48 ) } \end{array} \right. \Rightarrow \left\{ \begin{array} { l } { x + y = 48 } \\ { 0.80 x = 12 } \end{array} \right.\)

Use the second equation to find \(x\):

\(\begin{aligned} 0.80 x & = 12 \\ x & = \frac { 12 } { 0.80 } \\ x & = 15 \end{aligned}\)

\(\begin{aligned} x + y & = 48 \\ \color{OliveGreen}{15} + y & = 48 \\ y & = 33 \end{aligned}\)

We need to mix \(33\) liters of water with \(15\) liters of antifreeze concentrate.

Exercise \(\PageIndex{3}\)

A chemist wishes to create \(100\) ml of a solution with \(12\)% acid content. He uses two types of stock solutions, one with \(30\)% acid content and another with \(10\)% acid content. How much of each does he need?

The chemist will need to mix \(10\) ml of the \(30\)% acid solution with \(90\) ml of the \(10\)% acid solution.

www.youtube.com/v/NXbyJNE9mWw

Uniform Motion Problems (Distance Problems)

Recall that the distance traveled is equal to the average rate times the time traveled at that rate, \(D = r ⋅ t\). These uniform motion problems usually have a lot of data, so it helps to first organize that data in a chart and then set up a linear system. In this section, you are encouraged to use two variables.

Example \(\PageIndex{6}\):

An executive traveled a total of \(4\) hours and \(875\) miles by car and by plane. Driving to the airport by car, she averaged \(50\) miles per hour. In the air, the plane averaged \(320\) miles per hour. How long did it take her to drive to the airport?

We are asked to find the time it takes her to drive to the airport; this indicates that time is the unknown quantity.

Let \(x\) represent the time it took to drive to the airport. Let \(y\) represent the time spent in the air.

Fill in the chart with the given information.

Use the formula \(D = r \cdot t\) to fill in the unknown distances.

\(\begin{array} { l } { \text { Distance traveled in the car: } D = r \cdot t = 50 \cdot x } \\ { \text { Distance traveled in the air: } D = r \cdot t = 320 \cdot y } \end{array}\)

The distance column and the time column of the chart help us to set up the following linear system.

\(\left\{ \begin{array} { c } { x + y = \:4 \:\:\color{Cerulean}{\leftarrow total \:time\:traveled }} \\ { 50 x + 320 y = 875 \color{Cerulean}{\leftarrow total\: distance\: traveled } } \end{array} \right.\)

\(\begin{aligned} \color{red}{- 50 x}\color{black}{ -} 50 y& = - 200 \\ \pm\:\:\color{red}{ 50 x}\color{black}{ +} 320 y& = 875 \\ \hline\\270y&=675\\y&=\frac{675}{270}\\y&=\frac{5}{2} \end{aligned}\)

Now back substitute to find the time \(x\) it took to drive to the airport:

\(\begin{aligned} x + y & = 4 \\ x + \color{OliveGreen}{\frac { 5 } { 2 }} & \color{Black}{=} 4 \\ x & = \frac { 8 } { 2 } - \frac { 5 } { 2 } \\ x & = \frac { 3 } { 2 } \end{aligned}\)

It took her \(1 \frac{1}{2}\) hours to drive to the airport.

It is not always the case that time is the unknown quantity. Read the problem carefully and identify what you are asked to find; this defines your variables.

Example \(\PageIndex{7}\):

Flying with the wind, a light aircraft traveled \(240\) miles in \(2\) hours. The aircraft then turned against the wind and traveled another \(135\) miles in \(1 \frac{1}{2}\) hours. Find the speed of the airplane and the speed of the wind.

Begin by identifying variables.

Let \(x\) represent the speed of the airplane.

Let \(w\) represent the speed of the wind.

Use the following chart to organize the data:

With the wind, the airplane’s total speed is \(x + w\). Flying against the wind, the total speed is \(x − w\).

Use the rows of the chart along with the formula \(D = r ⋅ t\) to construct a linear system that models this problem. Take care to group the quantities that represent the rate in parentheses.

\(\left\{ \begin{array} { l } { 240 = ( x + w ) \cdot 2 \color{Cerulean}{\leftarrow distance\: traveled\: with \:the \:wind } } \\ { 135 = ( x - w ) \cdot 1.5 \color{Cerulean}{ \leftarrow distance\: traveled\: against\: the\: wind } } \end{array} \right.\)

If we divide both sides of the first equation by \(2\) and both sides of the second equation by \(1.5\), then we obtain the following equivalent system:

\(\left\{ \begin{array} { l } { 240 = ( x + w ) \cdot 2 } \quad\quad\overset{\div 2}{\Longrightarrow} \\ { 135 = ( x - w ) \cdot 1.5 \:\quad\underset{\div 1.5}{\Longrightarrow} } \end{array} \right. \quad \left\{ \begin{array} { l } { 120 = x + w } \\ { 90 = x - w } \end{array} \right.\)

Here \(w\) is lined up to eliminate.

\(\begin{aligned} x \color{red}{+ w}&\color{black}{ =} 120 \\ \pm x \color{red}{- w}&\color{black}{=} 90 \\ \hline\\2x &=210\\x & = \frac{210}{2} \\x& = 105\end{aligned}\)

Back substitute

\(\begin{aligned} x + w & = 120 \\ \color{OliveGreen}{105}\color{black}{ +} w & = 120 \\ w & = 15 \end{aligned}\)

The speed of the airplane is \(105\) miles per hour and the speed of the wind is \(15\) miles per hour.

Exercise \(\PageIndex{4}\)

A boat traveled \(27\) miles downstream in \(2\) hours. On the return trip, which was against the current, the boat was only able to travel \(21\) miles in \(2\) hours. What were the speeds of the boat and of the current?

The speed of the boat was \(12\) miles per hour and the speed of the current was \(1.5\) miles per hour.

www.youtube.com/v/EvdJQTFSUSs

Key Takeaways

• Use two variables as a means to simplify the algebraic setup of applications where the relationship between unknowns is unclear.
• Carefully read the problem several times. If two variables are used, then remember that you need to set up two linear equations in order to solve the problem.
• Be sure to answer the question in sentence form and include the correct units for the answer.

Exercise \(\PageIndex{5}\)

Set up a linear system and solve.

• The sum of two integers is \(45\). The larger integer is \(3\) less than twice the smaller. Find the two integers.
• The sum of two integers is \(126\). The larger is \(18\) less than \(5\) times the smaller. Find the two integers.
• The sum of two integers is \(41\). When \(3\) times the smaller is subtracted from the larger the result is \(17\). Find the two integers.
• The sum of two integers is \(46\). When the larger is subtracted from twice the smaller the result is \(2\). Find the two integers.
• The difference of two integers is \(11\). When twice the larger is subtracted from \(3\) times the smaller, the result is \(3\). Find the integers.
• The difference of two integers is \(6\). The sum of twice the smaller and the larger is \(72\). Find the integers.
• The sum of \(3\) times a larger integer and \(2\) times a smaller is \(15\). When \(3\) times the smaller integer is subtracted from twice the larger, the result is \(23\). Find the integers.
• The sum of twice a larger integer and \(3\) times a smaller is \(10\). When the \(4\) times the smaller integer is added to the larger, the result is \(0\). Find the integers.
• The difference of twice a smaller integer and \(7\) times a larger is \(4\). When \(5\) times the larger integer is subtracted from \(3\) times the smaller, the result is \(−5\). Find the integers.
• The difference of a smaller integer and twice a larger is \(0\). When \(3\) times the larger integer is subtracted from \(2\) times the smaller, the result is \(−5\). Find the integers.
• The length of a rectangle is \(5\) more than twice its width. If the perimeter measures \(46\) meters, then find the dimensions of the rectangle.
• The width of a rectangle is \(2\) centimeters less than one-half its length. If the perimeter measures \(62\) centimeters, then find the dimensions of the rectangle.
• A partitioned rectangular pen next to a river is constructed with a total \(136\) feet of fencing (see illustration). If the outer fencing measures \(114\) feet, then find the dimensions of the pen.

14. A partitioned rectangular pen is constructed with a total \(168\) feet of fencing (see illustration). If the perimeter measures \(138\) feet, then find the dimensions of the pen.

15. Find \(a\) and \(b\) such that the system \(\left\{ \begin{array} { l } { a x + b y = 8 } \\ { b x + a y = 7 } \end{array} \right.\) has solution \((2,1)\). (Hint: Substitute the given \(x\)- and \(y\)-values and solve the resulting linear system in terms of \(a\) and \(b\).)

16. Find \(a\) and \(b\) such that the system \(\left\{ \begin{array} { l } { a x - b y = 11 } \\ { b x + a y = 13 } \end{array} \right.\) has solution \((3, -1)\).

17. A line passes through two points \((5, −9)\) and \((−3, 7)\). Use these points and \(y = mx + b\) to construct a system of two linear equations in terms of \(m\) and \(b\) and solve it.

18. A line passes through two points \((2, 7)\) and \((\frac{1}{2}, −2)\). Use these points and \(y = mx + b\) to construct a system of two linear equations in terms of \(m\) and \(b\) and solve it.

19. A \($5,200\) principal is invested in two accounts, one earning \(3\)% interest and another earning \(6\)% interest. If the total interest for the year is \($210\), then how much is invested in each account?

20. Harry’s \($2,200\) savings is in two accounts. One account earns \(2\)% annual interest and the other earns \(4\)%. His total interest for the year is \($69\). How much does he have in each account?

21. Janine has two savings accounts totaling \($6,500\). One account earns \(2 \frac{3}{4}\)% annual interest and the other earns \(3 \frac{1}{2}\)%. If her total interest for the year is \($211\), then how much is in each account?

22. Margaret has her total savings of \($24,200\) in two different CD accounts. One CD earns \(4.6\)% interest and another earns \(3.4\)% interest. If her total interest for the year is \($1,007.60\), then how much does she have in each CD account?

23. Last year Mandy earned twice as much interest in her Money Market fund as she did in her regular savings account. The total interest from the two accounts was \($246\). How much interest did she earn in each account?

24. A small business invested \($120,000\) in two accounts. The account earning \(4\)% annual interest yielded twice as much interest as the account earning \(3\)% annual interest. How much was invested in each account?

25. Sally earns \($1,000\) per month plus a commission of \(2\)% of sales. Jane earns \($200\) per month plus \(6\)% of her sales. At what monthly sales figure will both Sally and Jane earn the same amount of pay?

26. The cost of producing specialty book shelves includes an initial set-up fee of \($1,200\) plus an additional \($20\) per unit produced. Each shelf can be sold for \($60\) per unit. Find the number of units that must be produced and sold where the costs equal the revenue generated.

27. Jim was able to purchase a pizza for \($12.35\) with quarters and dimes. If he uses \(71\) coins to buy the pizza, then how many of each did he have?

28. A cash register contains \($5\) bills and \($10\) bills with a total value of \($350\). If there are \(46\) bills total, then how many of each does the register contain?

29. Two families bought tickets for the home basketball game. One family ordered \(2\) adult tickets and \(4\) children’s tickets for a total of \($36.00\). Another family ordered \(3\) adult tickets and \(2\) children’s tickets for a total of \($32.00\). How much did each ticket cost?

30. Two friends found shirts and shorts on sale at a flea market. One bought \(4\) shirts and \(2\) shorts for a total of \($28.00\). The other bought \(3\) shirts and \(3\) shorts for a total of \($30.75\). How much was each shirt and each pair of shorts?

31. A community theater sold \(140\) tickets to the evening musical for a total of \($1,540\). Each adult ticket was sold for \($12\) and each child ticket was sold for \($8\). How many adult tickets were sold?

32. The campus bookstore sells graphing calculators for \($110\) and scientific calculators for \($16\). On the first day of classes \(50\) calculators were sold for a total of \($1,646\). How many of each were sold?

33. A jar consisting of only nickels and quarters contains \(70\) coins. If the total value is \($9.10\), how many of each coin are in the jar?

34. Jill has \($9.20\) worth of dimes and quarters. If there are \(68\) coins in total, how many of each does she have?

1. The integers are \(16\) and \(29\).

3. The integers are \(6\) and \(35\).

5. The integers are \(25\) and \(36\).

7. The integers are \(−3\) and \(7\).

9. The integers are \(−5\) and \(−2\).

11. Length: \(17\) meters; width: \(6\) meters

13. Width: \(22\) feet; length: \(70\) feet

15. \(a = 3, b = 2\)

17. \(m = −2, b = 1\)

19. \($3,400\) at \(3\)% and \($1,800\) at \(6\)%

21. \($2,200\) at \(2 \frac{3}{4}\)% and \($4,300\) at \(3 \frac{1}{2}\)%

23. Savings: \($82\); Money Market: \($164\).

25. \($20,000\)

27. \(35\) quarters and \(36\) dimes

29. Adults \($7.00\) each and children \($5.50\) each.

31. \(105\) adult tickets were sold.

33. The jar contains \(42\) nickels and \(28\) quarters.

Exercise \(\PageIndex{6}\)

• A \(17\)% acid solution is to be mixed with a \(9\)% acid solution to produce \(8\) gallons of a \(10\)% acid solution. How much of each is needed?
• A nurse wishes to obtain \(28\) ounces of a \(1.5\)% saline solution. How much of a \(1\)% saline solution must she mix with a \(4.5\)% saline solution to achieve the desired mixture?
• A customer ordered \(4\) pounds of a mixed peanut product containing \(12\)% cashews. The inventory consists of only two mixes containing \(10\)% and \(26\)% cashews. How much of each type must be mixed to fill the order?
• One alcohol solution contains \(10\)% alcohol and another contains \(25\)% alcohol. How much of each should be mixed together to obtain \(2\) gallons of a \(13.75\)% alcohol solution?
• How much cleaning fluid concentrate, with \(60\)% alcohol content, must be mixed with water to obtain a \(24\)-ounce mixture with \(15\)% alcohol content?
• How many pounds of pure peanuts must be combined with a \(20\)% peanut mix to produce \(2\) pounds of a \(50\)% peanut mix?
• A \(50\)% fruit juice concentrate can be purchased wholesale. Best taste is achieved when water is mixed with the concentrate in such a way as to obtain a \(15\)% fruit juice mixture. How much water and concentrate is needed to make a \(60\)-ounce fruit juice drink?
• Pure sugar is to be mixed with a fruit salad containing \(10\)% sugar to produce \(65\) ounces of a salad containing \(18\)% sugar. How much pure sugar is required?
• A custom aluminum alloy is created by mixing \(150\) grams of a \(15\)% aluminum alloy and \(350\) grams of a \(55\)% aluminum alloy. What percentage of aluminum is in the resulting mixture?
• A research assistant mixed \(500\) milliliters of a solution that contained a \(12\)% acid with \(300\) milliliters of water. What percentage of acid is in the resulting solution?

1. \(7\) gallons of the \(9\)% acid solution and \(1\) gallon of the \(17\)% acid solution

3. \(3.5\) pounds of the \(10\)% cashew mix and \(0.5\) pounds of the \(26\)% cashew mix

5. \(6\) ounces of cleaning fluid concentrate

7. \(18\) ounces of fruit juice concentrate and \(42\) ounces of water

Exercise \(\PageIndex{7}\)

• The two legs of a \(432\)-mile trip took \(8\) hours. The average speed for the first leg of the trip was \(52\) miles per hour and the average speed for the second leg of the trip was \(60\) miles per hour. How long did each leg of the trip take?
• Jerry took two buses on the \(265\)-mile trip from Los Angeles to Las Vegas. The first bus averaged \(55\) miles per hour and the second bus was able to average \(50\) miles per hour. If the total trip took \(5\) hours, then how long was spent in each bus?
• An executive was able to average \(48\) miles per hour to the airport in her car and then board an airplane that averaged \(210\) miles per hour. The \(549\)-mile business trip took \(3\) hours. How long did it take her to drive to the airport?
• Joe spends \(1\) hour each morning exercising by jogging and then cycling for a total of \(15\) miles. He is able to average \(6\) miles per hour jogging and \(18\) miles per hour cycling. How long does he spend jogging each morning?
• Swimming with the current Jack can swim \(2.5\) miles in \(\frac{1}{2}\) hour. Swimming back, against the same current, he can only swim \(2\) miles in the same amount of time. How fast is the current?
• A light aircraft flying with the wind can travel \(180\) miles in \(1 \frac{1}{2}\) hours. The aircraft can fly the same distance against the wind in \(2\) hours. Find the speed of the wind.
• A light airplane flying with the wind can travel \(600\) miles in \(4\) hours. On the return trip, against the wind, it will take \(5\) hours. What are the speeds of the airplane and of the wind?
• A boat can travel \(15\) miles with the current downstream in \(1 \frac{1}{4}\) hours. Returning upstream against the current, the boat can only travel \(8 \frac{3}{4}\) miles in the same amount of time. Find the speed of the current.
• Mary jogged the trail from her car to the cabin at the rate of \(6\) miles per hour. She then walked back to her car at a rate of \(4\) miles per hour. If the entire trip took \(1\) hour, then how long did it take her to walk back to her car?
• Two trains leave the station traveling in opposite directions. One train is \(8\) miles per hour faster than the other and in \(2 \frac{1}{2}\) hours they are \(230\) miles apart. Determine the average speed of each train.
• Two trains leave the station traveling in opposite directions. One train is \(12\) miles per hour faster than the other and in \(3\) hours they are \(300\) miles apart. Determine the average speed of each train.
• A jogger can sustain an average running rate of \(8\) miles per hour to his destination and \(6\) miles an hour on the return trip. Find the total distance the jogger ran if the total time running was \(1 \frac{3}{4}\) hour.

1. The first leg of the trip took \(6\) hours and the second leg took \(2\) hours.

3. It took her \(\frac{1}{2}\) hour to drive to the airport.

5. \(0.5\) miles per hour.

7. Airplane: \(135\) miles per hour; wind: \(15\) miles per hour

9. \(\frac{3}{5}\) hour

11. One train averaged \(44\) miles per hour and the other averaged \(56\) miles per hour.

Exercise \(\PageIndex{8}\)

• Compose a number or money problem of your own and share it on the discussion board.
• Compose a mixture problem of your own and share it on the discussion board.
• Compose a uniform motion problem of your own and share it on the discussion board.

1. Answer may vary

3. Answer may vary